Finding Horizontal Tangent Lines of an Equation

[Bo_]

The problem is to find the horizontal tangent lines of an equation. Here's my attempted differentiation.

y^2 = x^3 - x + 1

{dy/dx} = (3x^2 - 1)/(2y)

Correct, or no?

i'm going to need more help going forward even if that is right, I just want to make sure it is.

 

[montoyas7940]

Bo, I am pretty rusty at much of this but I will try to help since the forum is so empty at the moment.

It looks like you got the differentiation correct.

 

[EngageEngage]

deffinitely

 

[Bo_]

ok thanks, so assuming it's right, do set equal to y, then zero? In other words:

0 = (3x^2 - 1) / 2

and then quadratic formula using that^^^^? (remember I'm trying to find all slope zero tangent lines of the original equation.) If my procedural thinking is correct, then I don't think I need any more help, thanks.

 

[RyanSchw]

I'm trying to find all slope zero tangent lines of the original equation

You have found the slope for any point of that function except where the slope is undefined or wherever the graph may cross itself. To find the slope you would simply plug in your x and y values. However as you said you want to know the horizontal tangents.

You can't just simply set the y values to zero, you need to set the whole derivative to zero and solve the numerator for horizontal tangents, the denominator for vertical tangents.

 

[Bo_]

I can see clearly now the rain is gone, thanks.